Problem 29 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} $$

Step-by-Step Solution

Answer

The series diverges because the common ratio \( |r| > 1 \).

Step 1: Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \). Notice that it is a geometric series since each term can be written in the form \( ar^n \), where \( a = \frac{1}{\ln 2} \) and \( r = \frac{1}{\ln 2} \).

Step 2: Find the Common Ratio

Observe that each term is formed by multiplying the previous term by \( \frac{1}{\ln 2} \). Therefore, the common ratio \( r = \frac{1}{\ln 2} \). Since \( \ln 2 \approx 0.693 < 1 \), it follows that \( \frac{1}{\ln 2} > 1 \).

Step 3: Apply the Geometric Series Test

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the common ratio \( |r| < 1 \). Since \( \left| \frac{1}{\ln 2} \right| > 1 \), the series diverges by the geometric series test.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term is obtained by multiplying the previous term by a constant value, known as the common ratio. This type of series can be written in the form \( \sum_{n=0}^{\infty} ar^n \), where:

• \( a \) represents the first term of the series.
• \( r \) stands for the common ratio.

Such series are vital in mathematics because they can be used to model exponential growth or decay. Understanding geometric series is fundamental as they appear in diverse mathematical and real-world applications such as physics, finance, and computer science.
When working with geometric series, identifying the first term and the common ratio is crucial as these parameters determine the behavior of the series as \( n \) approaches infinity. Geometric series can either converge or diverge based on these values.

Common Ratio

The common ratio, denoted by \( r \), is the factor that determines how each term in a geometric series relates to the previous term. Simply put, it's the multiplier that gets you from one term to the next. If you have a term \( a \), the next term would be \( ar \), then \( ar^2 \), and so forth.

• If \( r \) is greater than one, the terms will grow larger, leading to divergence.
• If \( r \) is between zero and one, the terms will become smaller, often leading to convergence.
• If \( |r| \) equals one, the series neither converges nor diverges in a manner that is easy to determine by this factor alone.

In the provided example, \( \frac{1}{\ln 2} \) serves as the common ratio, which is found to be greater than one. This signifies that the series terms increase, resulting in divergence.

Geometric Series Test

The geometric series test is an effective method to determine whether a geometric series converges or diverges. According to this test, a geometric series \( \sum_{n=0}^{\infty} ar^n \) will converge if the absolute value of the common ratio \( |r| \) is less than one:

• If \( |r| < 1 \), the series converges.
• If \( |r| > 1 \), the series diverges.
• If \( |r| = 1 \), the test does not apply, and other methods may be necessary to determine convergence or divergence.

In our example with the series \( \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^{n}} \), the common ratio \( r \) is \( \frac{1}{\ln 2} \), which is greater than one. Applying the geometric series test here shows that because \(|r| > 1\), the series diverges. This test is a quick and reliable way to assess the behavior of a geometric series based solely on the common ratio.